CHAPTER 13
Simple Harmonic Motion
1
2
3
Oscillations
An object oscillates when it moves back and forth repeatedly, on either
side of some equilibrium position. If the object is stopped from
oscillating, it returns to the equilibrium position.
Oscillations can be observed in different ways:
1) Time keeping
2) Planetary Motion
3) Musical Instruments
4
Source: flickr
Source: Pixabay
Source: Pixabay
Natural (free) Oscillations
The easiest oscillations to understand are natural
oscillations, which every oscillator possesses. Every
oscillator has a natural frequency 𝒇𝟎 of vibration, the
frequency with which it vibrates freely after an initial
disturbance.
Simplest examples of natural oscillations: tuning fork and
pendulum.
5https://commons.wikimedia.org/wiki/File:Newtons_cradle.gif
Forced Oscillations
If a force is applied continuously to an oscillator, this will
cause the oscillator to vibrate in a forced oscillation. The
frequency of a forced oscillation is not the natural frequency
of the oscillator but rather the forcing frequency of the
external force.
6
Simple Harmonic Motion
A simple harmonic motion is a special kind of oscillations.
Definition:
A simple harmonic motion is defined as the motion of a
particle about a fixed point such that its acceleration 𝑎 is
proportional to its displacement 𝑥 from the fixed point, and
is directed towards the point.
7
Simple Harmonic Oscillators
1) Mass – Spring System
2) A Pendulum
8Wikipedia Commons
Source: Wikipedia Commons
Amplitude, Period and Frequency
The displacement changes between positive
and negative values.
The maximum displacement is called Amplitude, 𝒙𝟎 or A.
The period, T is the time for one complete oscillation.
The frequency,f is the number of oscillations per unit time,
a reciprocal of 𝑇.
𝑂: Equilibrium Position
𝑋, 𝑌:Extreme Points
9
X
Y
Amplitude, Period and Frequency
Recall:
Frequency, 𝑓 =1
𝑇
Angular frequency,
𝜔 = 2𝜋𝑓Or
𝜔 =2𝜋
𝑇
10
Example 1
For the graph below, determine the amplitude, frequency and time
period of the oscillation.
Answer:
Amplitude, 𝐴 = 0.8 m
Time period, 𝑇 = 3 × 2= 6 s
Frequency, 𝑓 =1
6
= 1.67 Hz
11
Simple Harmonic Motion
A system will oscillate if
there is a force acting on it
that tends to pull it back to
its equilibrium position – a
restoring force.
In a swinging pendulum the
combination of gravity and
the tension in the string that
always act to bring the
pendulum back to the centre
of its swing.
Tension
Gravity
Resultant
(restoring
force)
For a spring and mass the
combination of the gravity
acting on the mass and
the tension in the spring
means that the system will
always try to return to its
equilibrium position. M
T
g
Restoring
force
Simple Harmonic Motion
kxF
When the force acting on the particle is directly proportional
to the displacement and opposite in direction, the motion is
said to be simple harmonic motion.
xF
Simple Harmonic Motion
Relation of SHM with Circular Motion
The pattern of SHM is the same as the side view of an
object moving at a constant speed around a circular path
The amplitude of the
oscillation is equal to
the radius of the circle.
The object moves quickest
as it passes through the
central equilibrium position.
The time period of the
oscillation is equal to the
time taken for the object to
complete the circular path.Source: Wikimedia Commons
General SHM Equations
Most of the time, simple harmonic motion can be described by these two
basic trigonometric functions:
Where (𝜔𝑡 + 𝜙) denotes the phase of SHM. 𝜔 denotes the angular
frequency and 𝜙 is the phase constant.
𝑥 = 𝑥0 sin(𝜔𝑡 + 𝜙) or 𝑥 = 𝑥0 cos(𝜔𝑡 + 𝜙)
16
Phase Difference
The term phase describes the point that an oscillating mass has
reached within the complete cycle of an oscillation.
It is often important to describe the phase difference between two
oscillations.
17
Calculating Phase Difference
To calculate the phase difference, follow these steps:
1) Measure the time interval 𝑡 between two corresponding points on the
graphs.
2) Determine the period 𝑇 for one complete oscillation.
3) Calculate the phase difference as a fraction of an oscillation.
4) Convert to degrees or radians.
18
Question 2A body oscillating with SHM has a period of 1.5s and amplitude of 5cm. Calculate its frequency and maximum acceleration.
(a) f = 1 / T = 1 / 1.5sfrequency = 0.67 Hz
(b) a = - (2πf )2 xmaximum acceleration is when x = A (the amplitude)a = - (2πf )2 A
= - (2π x 0.6667Hz)2 x 0.015mmaximum acceleration = 0.88 ms-2
Question 3:Calculate the phase difference
between the two oscillations in
degrees and in radians.
Answer:
The time interval, 𝑡, is 17 ms.
Period, 𝑇 = 60 ms.
Phase difference,
𝜙 =𝑡
𝑇=
17
60= 0.283 oscillation
Phase difference in degree:
𝜙 = 0.283 × 360° = 102°
Phase difference in radian:
𝜙 = 0.283 × 2𝜋 = 1.78 rad
20
Question 4
The motion of a body in simple harmonic motion is described by the
equation:
𝑥 = 4.0 cos(2𝜋𝑡 +𝜋
3)
Where 𝑥 is in metres, 𝑡 is in seconds and (2𝜋𝑡 +𝜋
3) is in radians.
(a) What is (i) the displacement, (ii) the velocity, (iii) the acceleration
and (iv) the phase when 𝑡 = 2.0 s?
(b) Find (i) the frequency and (ii) the period of the motion.
21
22
Solution:
(a) (i) Displacement, 𝑥 = 4.0 cos(2𝜋𝑡 +𝜋
3)
when, 𝑡 = 2.0 s,
𝑥 = 4.0 cos(4𝜋 +𝜋
3)
=2.0 m.
(ii) Velocity, 𝑣 =𝑑𝑥
𝑑𝑡= −2𝜋 4.0 sin(2𝜋𝑡 + 𝜋/3)
When 𝑡 = 2.0s,
𝑣 = −8.0𝜋 sin(4𝜋 +𝜋
3)
= −21.8 ms-1
Solution (continued):
(iii)Acceleration, 𝑎 =𝑑𝑣
𝑑𝑡= − 4𝜋2 4.0 cos(2𝜋𝑡 +
𝜋
3)
When 𝑡 = 2.0 s,
𝑎 = − 4𝜋2 4.0 cos 4𝜋 +𝜋
3= −79.0 ms-2.
(iv) Phase = (2𝜋𝑡 +𝜋
3)
When t=2.0s, phase = (4𝜋 +𝜋
3) rad = 13.6 rad.
(b) (i) 𝜔 = 2𝜋, 𝑓 =𝜔
2𝜋=
2𝜋
2𝜋= 1 Hz
(ii) 𝑇 =1
𝑓= 1s
23
Simple Harmonic Motion
Consider a simple harmonic oscillation, described by the formula
𝑥 = 𝑥0 cos(𝜔𝑡)
The velocity of its vibration is described as
𝑣 =𝑑𝑥
𝑑𝑡= −𝜔𝑥0 sin𝜔𝑡
The acceleration
𝑎 =𝑑𝑣
𝑑𝑡= −𝜔2𝑥0 cos𝜔𝑡
But 𝑥0 cos𝜔𝑡 = 𝑥.
Hence,
𝑎 = −𝜔2𝑥.
24
Acceleration in SHM
25
x
a
+ A- A
- ω2 x
+ ω2 x
gradient = - ω2 or - (2πf )2
Simple Harmonic Motion
𝑑2𝑥
𝑑𝑡2= −𝜔2𝑥
𝑣𝑑𝑣
𝑑𝑥= −𝜔2𝑥
By separation of variables,
𝑣𝑑𝑣 = −𝜔2𝑥𝑑𝑥Integrating them,
𝑣2
2= −
𝜔2𝑥2
2+ 𝑐
When 𝑥 = ±𝑥0, 𝑣 = 0.
𝑐 =𝜔2𝑥0
2
2Hence, 𝑣2 = 𝜔2 𝑥0
2 − 𝑥2
Or
When v= max,
𝑣 = ±𝜔 𝑥02 − 𝑥2
26𝑣0 = ±𝜔𝑥0
Simple Harmonic Motion
𝑂
A B
-𝑥 +𝑥
+𝑎 −𝑎
𝑥 = −𝑥0𝑣 = 0𝑑2𝑥
𝑑𝑡2= 𝜔2|𝑥|
Max. PE
Zero KE
𝑥 = 0𝑣 = ±𝜔𝑎𝑑2𝑥
𝑑𝑡2= 0
Zero PE
Max KE
𝑥 = +𝑥0𝑣 = 0𝑑2𝑥
𝑑𝑡2= −𝜔2|𝑥|
Max. PE
Zero KE
27
Example:
28
Answer:(a) Amplitude =0.15 m
(b) Period 𝑇 = 2.0s
(c) Frequency 𝑓 =1
𝑇=
1
2.0= 0.5 Hz
(d) Angular velocity 𝜔 = 2𝜋𝑓 = 3.14 rad s−1
(e) (i) When 𝑥 = 0,𝑑2𝑥
𝑑𝑡2= 𝜔2 0 = 0
(ii) When 𝑥 = 0.15 m, |𝑑2𝑥
𝑑𝑡2|=𝜔2𝑥 = 3.14 2(0.15)
(f) Maximum velocity occurs when 𝑥 = 0, i.e. 𝑣𝑚𝑎𝑥 =𝜔𝑥0 = 3.14 0.15 = 0.47ms-1
Note: always use radian in calculations involving SHM unless otherwise
required by the questions.
29
The Period of a Simple Spring-Mass System
Consider a spring suspended with a mass 𝑚. If the elastic limit is not
exceeded,
𝐹 = 𝑘𝑒 = 𝑚𝑔where 𝑘 is the spring constant.
If the mass is pulled down and released, it oscillates with simple harmonic
motion. The tension 𝐹′ is equal to 𝑘(𝑒 + 𝑥).The restoring force on the mass is given by
𝑚𝑑2𝑥
𝑑𝑡2= 𝑚𝑔 − 𝐹′ = 𝑚𝑔 − 𝑘(𝑒 + 𝑥)
But 𝑘𝑒 = 𝑚𝑔, hence
𝑚𝑑2𝑥
𝑑𝑡2= −𝑘𝑥
𝑑2𝑥
𝑑𝑡2= −
𝑘
𝑚𝑥
𝜔 =𝑘
𝑚⇒ T = 2𝜋
𝑚
𝑘30
The Period of a Simple Pendulum
Consider a simple pendulum:
𝐹𝑟𝑒𝑠 = 𝑚𝑎𝑡𝑎𝑛𝑚𝑔 sin 𝜃 = −𝑘𝑠= −𝑘(𝑙𝜃)
𝑘 =𝑚𝑔 sin 𝜃
𝑙𝜃
Small angle approximation:
lim𝑥→0
sin 𝜃
𝜃= 1 ;
Equation becomes: 𝑘 =𝑚𝑔
𝑙; Substitute into T = 2𝜋
𝑚
𝑘:
𝜔 =𝑔
𝑙
𝑇 = 2𝜋𝑙
𝑔Christiaan Huygen’s
law
31
Small Angle ApproximationHowever, for small angles, sin θ and θ are approximately equal.
Energy in Simple Harmonic Motion
Kinetic Energy:
𝐸𝑘 =1
2𝑚𝑣2
But 𝑣 = ±𝜔 𝑥02 − 𝑥2
⇒ 𝐸𝑘 =1
2𝑚𝜔2(𝑥0
2 − 𝑥2)
Potential Energy:
𝐹𝑟𝑒𝑠 = −𝑚𝜔2𝑥Potential Energy = average restoring force × displacement
𝐸𝑃 =1
2𝑚𝜔2𝑥2
33
Total Energy in Simple Harmonic Motion
Consider the oscillation of a spring,
The potential energy of a spring is given by:
𝑈 =1
2𝑘𝑥2
=1
2𝑘𝑥0
2 sin2 𝜔𝑡 + 𝜙
The kinetic energy of the mass is given by:
𝐾 =1
2𝑚𝑣2
=1
2𝑚𝑥0
2𝜔2 cos2(𝜔𝑡 + 𝜙)
=1
2𝑘𝑥0
2 cos2(𝜔𝑡 + 𝜙)
The total mechanical energy,
𝐸 = 𝑈 + 𝐾 =1
2𝑘𝑥0
2
OR 𝐸 =1
2𝑚𝜔2 𝑥0
2 − 𝑥2 +1
2𝑚𝜔2𝑥2
=1
2𝑚𝜔2𝑥0
2
The total mechanical energy is constant and is independent of the time.34
Free and Damped Oscillations
Recall:
A particle is said to be undergoing free oscillations when the
only external force acting on it is the restoring force.
In real situations, however, frictional and other resistive forces
cause the oscillator’s energy to be dissipated, and this energy is
converted eventually into thermal energy. The oscillations are
said to be damped.
A displacement-time graph of the system shows an exponential
decay of the amplitude with time.
35time
Energy Dissipated energy
36
Example:A particle of mass 0.40 kg oscillates in simple harmonic motion with
frequency 5.0 Hz and amplitude 12 cm. Calculate, for the particle at
displacement 10 cm:
(a) The kinetic energy
𝐸𝑘 =1
2𝑚𝜔2 𝑥0
2 − 𝑥2 =1
20.40 2𝜋 5.0
20.122 − 0.102
= 0.87 J
(b) The potential energy
𝐸𝑝 =1
2𝑚𝜔2𝑥2 =
1
20.40 2𝜋 5.0
2(0.102)
= 1.97J
(c) Total energy
𝐸𝑡𝑜𝑡 =1
2𝑚𝜔2𝑥0
2
=1
20.40 2𝜋 5.0
2(0.122)
= 2.84J
Damped Oscillations
(a) Light damping – The system oscillates about the equilibrium position with
decreasing amplitude over a period of time. (ex: a mass and spring in air)
(b) Critical damping – The system does not oscillate but returns to the
equilibrium position in the shortest time. (Ex: the suspension system of a
car)
(c) Heavy damping − the system does not oscillate and damping is so large
that it will take a long time to reach the equilibrium position. (example:
coiled spring mattress)
37
Forced Oscillations and Resonance
Resonance is a special case of the forced oscillation. If the forcing
frequency of an external force happens to match the natural frequency of
oscillation of the system, the amplitude of the resulting oscillations can build
up to become very large.
Barton’s pendulum is a demonstration of the resonance phenomenon.
Several pendulums of different lengths hang from a horizontal string. The
‘driver’ pendulum P at the right has a large mass. If it is set swinging, the
pendulum 𝑌 will oscillate with the largest amplitude due to the fact that its
natural frequency matches the natural frequency of the driver pendulum
(their lengths are the same).
38
Resonance
In resonance, energy is transferred from the driver to the resonating system
much more efficiently.
For any system in resonance:
a) Its natural frequency is equal to the frequency of the driver.
b) Its amplitude is maximum
c) It absorbs the greatest possible energy from the driver.
39
Resonance Curves
applied force frequency, fA
amplitude of driven
system, A
driving
force
amplitude
f0
more damping
light damping
very light damping
Notes on the resonance curves
If damping is increased then the amplitude of the driven system is decreased at all driving frequencies.
If damping is decreased then the sharpness of the peak amplitude part of the curve increases.
The amplitude of the driven system tends to be:- Equal to the driving system at very low frequencies.
- Zero at very high frequencies.- Infinity (or the maximum possible) when fA is equal to f0 as
damping is reduced to zero.
The Tacoma Narrows Bridge Collapse
An example of resonance
caused by wind flow.
Washington State USA,
November 7th 1940.
Source: Picryl
Review Questions1) The pendulum of a clock is displaced by a distance of 4.0 cm and it
oscillates in s.h.m. with a frequency of 1.0 s.
a) Write down an equation to describe the displacement 𝑥 of the pendulum
bob with time 𝑡.b) Calculate:
i) The maximum velocity of the pendulum bob.
ii) Its velocity when its displacement is 2.0 cm.
43
Review Questions2) (a) Describe the energy changes during one complete oscillation of an
undamped simple pendulum.
(b) Explain why a lightly damped simple pendulum experiences a slow
decrease in its amplitude.
44
Review Questions3) The simple harmonic motion of a particle is given by
𝑎 = −𝑏𝑥Where 𝑎 is the acceleration,
𝑥 is the displacement of the particle and
𝑏 is a constant.
Write an expression of the time period of oscillation in term of 𝑏.
45
Review Questions4) A 50 g mass is attached to a securely clamped spring. The mass is pulled
downwards by 16 mm and released which causes it to oscillate with s.h.m. of
time period of 0.84 s.
a) Calculate the frequency of the oscillation.
b) Calculate the maximum velocity of the mass.
c) Calculate the maximum kinetic energy of the mass and state at which
point in the oscillation it will have this velocity.
d) Write down the maximum gravitational potential energy of the mass
(relative to its equilibrium position). You may assume the damping is
negligible.
46
Review Question
5)
9702/41/M/J/13
47
Review Question 5(continued)
48
Review Question 5) (continued)
49
50
Extra References:
http://www.acoustics.salford.ac.uk/feschools/waves/shm3.php
http://www.physbot.co.uk/further-mechanics.html
http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html#c3
51
